TITLE: What are the closed points of $\mathbb{A}_{\mathbb{R}}^2 = \operatorname{Spec}(\mathbb{R}[x,y])$?
QUESTION [6 upvotes]: I am trying to find all the closed points of $\mathbb{A}_{\mathbb{R}}^2$.
After a quick google research, I found that $\mathbb{A}_{\mathbb{R}}^2 = \operatorname{Spec}(\mathbb{R}[x,y])$ and then all we need to find is the maximal ideals of $\mathbb{R}[x,y]$.
However, set-theoretically this equalty does not make sense, as we are talking about points and prime ideals.
And more than that, how do I find the maximal ideals of $\mathbb{R}[x,y]$? Seems hard.
Any help would be appreciated.
Thanks!

REPLY [3 votes]: Re: 

However, set-theoretically this equalty does not make sense, as we are talking about points and prime ideals.

The points of a scheme are*, by definition, prime ideals of a ring.  If this is new to you, you're probably not ready to be looking at this question until you go back and learn what a scheme is.
Once you understand the question, Slade's answer shows you a good way of working the problem.
*Or, if you want to be incredibly pedantic, correspond to the ideals of a ring, since a scheme is a locally ringed space that happens to be locally isomorphic to the spectrum of a ring.